Enter five million, twenty thousand, twenty six in expanded form

The table below represents what type of function?

Troyanec [42]

Answer:

nonlinear because the second row isn't going up by the same number

3 0

2 years ago

Which equation of a line passes through the points (3, −1) and (6, 1)?

Vaselesa [24]

Answer:

Y = 2/3x - 3

Step-by-step explanation:

Recall the general equation for a straight line is

y = mx + b

where m is the gradient and b is the y-intercept

given 2 points whose coordinates are (x1, y1) and (x2, y2), m can be found with the following formula:

m = $\frac{y1-y2}{x1-x2}$

in this case, x1 = 3, y1 = -1, x2 = 6, y2=1

applying these values to the formula for m will give

m = (-1 -1) / (3-6) = 2/3

We can see immediately that only the first (top-most) answer has this value for m and we can guess that this is probably the answer.

But we can still check to confirm:

If we substitute this back into the general equation, we get:

y = (2/3)x + b

In order to find the value for b, we substitute any one of the 2 given points back into this equation. Lets choose (6,1)

1 = (2/3)(6) + b

1 = 4 + b

b = -3

Confirm Y = 2/3x - 3 is the answer.

4 0

3 years ago

You are a landlord, and your tenant complains that you are in violation of the ADA because your wheelchair ramp is too steep. An

Savatey [412]

Answer:

Not justified

Step-by-step explanation:

ADA requirements are for a slope of 1 in 12 or less, an 8.3% grade, or an angle of 4.8° (actually, 4.764° when rounded to 3 decimal places). The ramp angle found by the assessor is compliant.

The tenant complaint does not seem justified.

3 0

3 years ago

What is the approximate volume of the

zhannawk [14.2K]

The answer is : 10416.67ft^3

8 0

3 years ago

Read 2 more answers

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

4 0

3 years ago

Read 2 more answers